POJ 2100 Graveyard Design 尺取法

                                                                                                   Graveyard Design

                                     Time Limit:10000MS      Memory Limit:64000KB      64bit IO Format:%lld & %llu  

     

Description

King George has recently decided that he would like to have a new design for the royal graveyard. The graveyard must consist of several sections, each of which must be a square of graves. All sections must have different number of graves.
After a consultation with his astrologer, King George decided that the lengths of section sides must be a sequence of successive positive integer numbers. A section with side length s contains s² graves. George has estimated the total number of graves that will be located on the graveyard and now wants to know all possible graveyard designs satisfying the condition. You were asked to find them.

Input

Input file contains n --- the number of graves to be located in the graveyard (1 <= n <= 10¹⁴ ).

Output

On the first line of the output file print k --- the number of possible graveyard designs. Next k lines must contain the descriptions of the graveyards. Each line must start with l --- the number of sections in the corresponding graveyard, followed by l integers --- the lengths of section sides (successive positive integer numbers). Output line's in descending order of l.

Sample Input

2030

Sample Output

24 21 22 23 243 25 26 27

题意：给你一个数 求一个连续序列的元素的平方之和 等于这个数 输出数量 还有区间长度及各个元素

ACcode：

#include<iostream>#include<cstdio>#include<vector>#include<cmath>#include<algorithm>using namespace std;typedef long long LL;typedef pair<LL, LL>pii;vector<pii>v;int main(){long long  n, k, ans = 0;while (scanf("%I64d",&n)!=EOF){k = sqrt(n); v.clear();LL l = 1, r = 1, cnt = 1;while (1){while (r <= k && cnt < n){r++;cnt += r*r;}if (cnt < n ) break;if (cnt == n){v.push_back(make_pair(l, r));}cnt -= l*l; l++;}LL len = v.size();printf("%I64d\n", len);for (int i = 0; i < len; i++){printf("%I64d", v[i].second - v[i].first + 1);for (LL j = v[i].first; j <= v[i].second ; j++){printf(" %I64d", j);}printf("\n");}}return 0;}